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3 years ago

14

Suppose you have an accounting background and you are considered an expert in the field of discovering errors in electronic spre

adsheets. You are considered a knowledge engineer if you are part of a team that is developing an expert system to discover errors in spreadsheets. True or false?

1 answer:

leonid [27]3 years ago

6 0

Answer: True

Explanation: A knowledge engineer is a professional who works to improve and builds with the aid of science and also implements advanced logic and AI( artificial intelligence) in computer systems in order to investigate particular issues.

A knowledge engineer also helps tries to emulate the judgements of man.

Since you are part of the team involved in discovering errors in the electronic spread sheet you are a knowledge engineer.

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What should a sailboat operator do when approaching a PWC head-on

nekit [7.7K]

<u><em>In accordance with the International Regulation for the prevention of collisions at sea</em></u><u>: </u>

<u>1.- A sailing boat has a passing preference over a motorized boat, </u><u>except when the motor boat is limited by its draft</u><u>. </u>

<u>2.- The sailboat must maintain its course and speed. </u>

<u>3.- </u><em><u>If it is evident that the PWC does not respond</u></em><u>, the sailboat must sound the warning signal, and change its course to starboard. </u>

<u>4.- </u><u><em>All actions must be taken as soon as possible</em></u><u>. </u>

<u>5.- If a sailboat is using its engine, the situation changes, and in that case, both ships must alter to starboard.</u>

8 0

3 years ago

At the beach in San Francisco (0 meters) the pressure of the atmosphere is 101.325 kPa

Korvikt [17]

Answer:

$P = -\frac{17978}{1609344}(h)+101.325$

Explanation:

Given

$h = height$

$P = Pressure$

$(h_1,P_1) = (0,101.325)$

$(h_2,P_2) = (1609.344 ,83.437 )$

Required

Determine the linear equation for P in terms of h

First, we calculate the slope/rate (m);

The following formula is used:

$m = \frac{P_2 - P_1}{h_2 - h_1}$

Substitute values for P's and h's

$m = \frac{83.347 - 101.325}{1609.344- 0}$

$m = \frac{-17.978}{1609.344}$

$m = -\frac{17.978}{1609.344}$

Multiply by 1000/1000

$m = -\frac{17.978 * 1000}{1609.344*1000}$

$m = -\frac{17978}{1609344}$

The equation is then calculated using:

$P - P_1 = m(h - h_1)$

Substitute values for m, h1 and P1

$P - P_1 = m(h - h_1)$

$P - 101.325 = -\frac{17978}{1609344}(h - 0)$

$P - 101.325 = -\frac{17978}{1609344}(h)$

Make P the subject

$P = -\frac{17978}{1609344}(h)+101.325$

<em>The above is the required linear equation</em>

8 0

3 years ago

When electrons are accelerated by 2450 V in an electron microscope, they will have

Anni [7]

I think the answer to this is 0.811nm

5 0

3 years ago

What is the mass of a rock lifted 2 meters off the ground that has 196 J of potential energy?

eimsori [14]

Answer:

10kg

Explanation:

Let PE=potential energy

PE=196J

g(gravitational force)=9.8m/s^2

h(change in height)=2m

m=?

PE=m*g*(change in h)

196=m*9.8*2

m=10kg

4 0

3 years ago

A car is stopped at a traffic light. When the light turns green at t=0, a truck with a constant speed passes the car with a 20m/

s344n2d4d5 [400]

Answer:

At $t = (70 / 3) \; {\rm s}$ (approximately $23.3 \; {\rm s}$ .)

Explanation:

Note that the acceleration of the car between $t = 0\; {\rm s}$ and $t = 20\; {\rm s}$ ( $\Delta t = 20\; {\rm s}$ ) is constant. Initial velocity of the car was $v_{0} = 0\; {\rm m\cdot s^{-1}}$ , whereas $v_{1} = 35\; {\rm m\cdot s^{-1}}$ at $t = 20\; {\rm s}\!$ . Hence, at $t = 20\; {\rm s}\!\!$ , this car would have travelled a distance of:

$\begin{aligned}x &= \frac{(v_{1} - v_{0})\, \Delta t}{2} \\ &= \frac{(35\; {\rm m\cdot s^{-1}} - 0\; {\rm m\cdot s^{-1}}) \times (20\; {\rm s})}{2} \\ &= 350\; {\rm m}\end{aligned}$ .

At $t = 20\; {\rm s}$ , the truck would have travelled a distance of $x = v\, t = 20\; {\rm m\cdot s^{-1}} \times 20\; {\rm s} = 400\; {\rm m}$ .

In other words, at $t = 20\; {\rm s}$ , the truck was $400\; {\rm m} - 350\; {\rm m} = 50\; {\rm m}$ ahead of the car. The velocity of the car is greater than that of the truck by $35\; {\rm m\cdot s^{-1}} - 20\; {\rm m\cdot s^{-1}} = 15 \; {\rm m\cdot s^{-1}}$ . It would take another $(50\; {\rm m}) / (15\; {\rm m\cdot s^{-1}}) = (10/3)\; {\rm s}$ before the car catches up with the truck.

Hence, the car would catch up with the truck at $t = (20 + (10/3))\; {\rm s} = (70 / 3)\; {\rm s}$ .

3 0

2 years ago